2Al + 3Cl₂ → 2AlCl₃
mol Al = 2/3 x 1.25 = 0.83
mass Al = 0.83 x 27 g/mol = 22.41 g
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Answer:
Answer is water, carbon dioxide, and energy, which yields glucose and oxygen, as in the well-known formula: 6 H2O + 6 CO2 + Energy = C6H12O6 + 6 O2. Hope it helps!
-173.15
-0.15
-267.15
416.15
846.15
assuming k is kelvins and c is celsius
A control group is the group in an experiment that does not receive any sort of change, to then be compared to the other treated objects at the end of the study.
